Question 53784
Let h=12 and solve for t:
12=8t-t^2  set equal to zero(add -8t and t^2 to both sides)
t^2-8t+12=0    can be factored:
(t-6)(t-2)=0    zero product rule:
t-6=0   and t-2=0
t=6     and t=2
The ball is at a height of 12 at times 2 seconds and 6 seconds.
The first time it reaches a height of 12 is 2 seconds.
check:
h=8(2)-2^2
h=16-4=12